Answer:
0.75M Fe²⁺
Explanation:
First, we need to balance the redox reaction in acidic medium. Then, we can obtain moles of KMnO4 and with the reaction moles and molarity of the Fe²⁺ solution:
<em>Redox Balance:</em>
Fe²⁺ → Fe³⁺ + 1e⁻
5e⁻ + 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O
___________________________
5Fe²⁺ + 5e⁻ + 8H⁺ + MnO₄⁻ → 5Fe³⁺ + 5e⁻ + Mn²⁺ + 4H₂O
<h3>5Fe²⁺ + 8H⁺ + MnO₄⁻ → 5Fe³⁺ + Mn²⁺ + 4H₂O</h3>
<em>Moles of KMnO₄:</em>
70.0mL = 0.0700L * (0.150mol / L) = 0.0105 moles KMnO₄
<em>Moles and molarity Fe²⁺:</em>
0.0105 moles KMnO₄ * (5 moles Fe²⁺ / 1mol KMnO₄) = 0.0525 moles Fe²⁺
In 70.0mL = 0.0700L:
0.0525 moles Fe²⁺ / 0.0700L =
<h3>0.75M Fe²⁺</h3>